unionall用法

Posted 2023-05-05

参考技术A union all用法举例如下：
SELECT * FROM
(SELECT top 1 left(convert(varchar(100),RECEIVE_TIME,108 ),5) receiveTime, \'0\' restoreTime,unit_code,unit_name
FROM T_FILE_RECEIVE_RECORD WHERE UNIT_CODE = \'123123\' and convert(char(100),RECEIVE_TIME,23)=\'2018-10-06\' ORDER BY RECEIVE_TIME DESC
) table1
UNION ALL
SELECT * FROM
(SELECT top 1 \'0\' receiveTime, left(convert(varchar(100),RESTORE_TIME,108 ),5) restoreTime,unit_code,unit_name
FROM T_FILE_RESTORE_RECORD WHERE UNIT_CODE = \'123123\' and convert(char(100),restore_time,23)=\'2018-10-06\' ORDER BY restore_time DESC
) table2

扩展资料

使用UNION(或者UNION ALL)语句时，如果UNION的\'两个结果集在单独排序后再拼接，则他们的ORDER BY是失效的。如果我们要进行排序有以下两种方法：
将它们作为子查询再ORDER BY查询一次；在第一个结果集中不使用排序，且不用括号分隔，而在第二个结果集后使用ORDER BY。

with recursive用法

with recursive 则是一个递归的查询子句，他会把查询出来的结果再次代入到查询子句中继续查询。

with recursive d(n, fact) as (
values (1,2)
union all #合并
select n+1, (n+1)*fact from d where n < 5)
SELECT * from d;

递归过程如下：
n=1 fact=2
n=1,n<5: n=1+1=2,fact=(1+1)*2=4
n=2,n<5:n=2+1=3,fact=(2+1)*4=12
n=3,n<5:n=3+1=4,fact=(3+1)*12=48
n=4,n<5:n=4+1=5,fact=(4+1)*48=240
n=5 n>=5==stop

with recursive d(n, fact) as (
values (1,2)
union all
select n+2, (n+1)*fact from d where n < 5)
SELECT * from d;

递归过程如下：
n=1 fact=2
n=1,n<5: n=1+2=3,fact=(1+1)*2=4
n=3,n<5:n=3+2=5,fact=(3+1)*4=16
n=5 n>=5==stop

with recursive d(n, fact) as (
values (1,2)
union all
select n+2, (n+1)*fact from d where n < 5)
select sum(fact) from d;

sum(fact)=2+4+16=22

with recursive d(n, fact) as (
values (1,2)
union all
select n+2, (n+1)*fact from d where n < 5)
select sum(n) from d;

sum(n)=1+3+5=9

select * from company;

with recursive t(n) as (
   values (10)
   union all
   select salary from company where salary < 20000
)
select  * from t;

with recursive t(n) as (
   values (10)
   union all
   select salary from company where salary < 20000
)
select  sum(n) from t;